268 lines
7.8 KiB
TeX
268 lines
7.8 KiB
TeX
\section{Half a Qubit}
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First, a pair of definitions. We've used both these terms implicitly in the previous section,
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but we'll need to introduce proper definitions before we continue.
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\definition{}
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A \textit{linear combination} of two vectors $u$ and $v$ is the sum $au + bv$ for scalars $a$ and $b$. \par
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\note[Note]{In other words, a linear combination is exactly what it sounds like.}
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\definition{}
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A \textit{normalized vector} (also called a \textit{unit vector}) is a vector with length 1.
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\vspace{4mm}
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\begin{tcolorbox}[
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enhanced,
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breakable,
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colback=white,
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colframe=ored,
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boxrule=0.6mm,
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arc=0mm,
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outer arc=0mm,
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]
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\color{ored}
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\begingroup
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\large\centering
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\textbf{Disclaimer:} \par
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\endgroup
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\vspace{1ex}
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The \say{qubits} we're about to define aren't \textit{really} qubits. The proper definition is a bit more
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complicated, but don't worry about that yet. For now, take what I say as truth---we'll get to
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the complex definition soon enough.
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\vspace{2mm}
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The information provided in this handout does not, and is not intended to, constitute legal advice.
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All information, content, and material in this document is for general informational purposes only.
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\end{tcolorbox}
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\generic{Remark:}
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Just like a classical bit, a \textit{quantum bit} (or \textit{qubit}) can take the values $\ket{0}$ and $\ket{1}$. \par
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However, \texttt{0} and \texttt{1} aren't the only states a qubit may have.
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\vspace{2mm}
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We'll make sense of quantum bits by extending the \say{vectored} bit representation we developed in the previous section.
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First, let's look at a diagram we drew a few pages ago:
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\begin{ORMCbox}{Time Travel (Page 2)}{black!10!white}{black!65!white}
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A classical bit takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par
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We'll represent \texttt{0} and \texttt{1} as perpendicular unit vectors $\ket{0}$ and $\ket{1}$,
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show below.
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\node[right] at (1.2, 0) {$\ket{0}$};
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below] at (1, 0) {\texttt{0}};
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\draw[->] (0, 0) -- (0, 1.2);
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\node[above] at (0, 1.2) {$\ket{1}$};
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[left] at (0, 1) {\texttt{1}};
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\end{tikzpicture}
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\end{center}
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The point marked $1$ is at $[0, 1]$. It is no parts $\ket{0}$, and all parts $\ket{1}$. \par
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Of course, we can say something similar about the point marked $0$: \par
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It is at $[1, 0] = (1 \times \ket{0}) + (0 \times \ket{1})$, and is thus all $\ket{0}$ and no $\ket{1}$. \par
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\end{ORMCbox}
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The diagram in the box above can also be used to describe the state of a qubit. \par
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Like classical bits, qubits have the \textit{basis states} $\ket{0}$ and $\ket{1}$. \par
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Unlike classical bits, qubits may take values that are some combination of both.
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\vspace{2mm}
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Namely, every possible state of a qubit is a \textit{normalized linear combination} of $\ket{0}$ and $\ket{1}$. \par
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Such states are called \textit{superpositions} of $\ket{0}$ and $\ket{1}$, since they partially contain both states.
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\vfill
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\pagebreak
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\definition{}
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The state of a quantum bit is the column unit vector
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$
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\ket{\psi}
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= \left[\begin{smallmatrix} a \\ b \end{smallmatrix}\right]
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= a\ket{0} + b\ket{1}$ for $a, b \in \mathbb{R}
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$. \par
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Note that the length of $\ket{\psi}$ must always be $1$, which is the same as saying that $a^2 + b^2 = 1$.
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\vspace{2mm}
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If we plot the set of valid quantum states on our plane, we get a unit circle centered at the origin: \par
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
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\node[above right] at (0.87, 0.5) {$\ket{\psi}$};
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\end{tikzpicture}
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\end{center}
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\problem{}
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In the above diagram, the counterclockwise angle from $\ket{0}$ to $\ket{\psi}$ is $30^\circ$\hspace{-1ex}. \par
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Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
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\vfill
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\definition{Measurement I}
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Although a qubit may have many states, it must be $\ket{0}$ or $\ket{1}$ when we measure it. \par
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\vspace{2mm}
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As a trivial example, say $\ket{\psi}$ = $\ket{0}$. \par
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If we were to measure $\ket{\psi}$, we'd get $\ket{0}$, and the state of the qubit wouldn't change.
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\vspace{2mm}
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However, something interesting happens when $\ket{\psi} = a\ket{0} + b\ket{1}$. \par
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Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following probabilities: \par
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\begin{itemize}[itemsep = 2mm, topsep = 2mm]
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\item $\mathcal{P}(\ket{1}) = a^2$
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\item $\mathcal{P}(\ket{0}) = b^2$
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\end{itemize}
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\note{
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Note that $\mathcal{P}(\ket{0}) + \mathcal{P}(\ket{1}) = 1$. \\
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As you already know, this is true of any probability function.
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}
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\vspace{2mm}
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In addition $\ket{\psi}$ \textit{collapses} to the state we measure: it instantly jumps to the state we measure, \par
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leaving no trace of its previous state. If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
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it will remain in that state until it is changed.
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\vspace{2mm}
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Quantum bits \textit{cannot} be measured without their state collapsing. \par
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We cannot certainly know the state of a qubit unless that state is $\ket{0}$ or $\ket{1}$.
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\pagebreak
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\problem{}
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\begin{itemize}
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\item What is the probability we get $\ket{0}$ if we measure $\ket{\psi_0}$? \par
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\item What outcomes can we get if we measure it a second time? \par
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\item What are these probabilities for $\ket{\psi_1}$?
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\end{itemize}
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\draw[dotted] (0, 0) -- (0.87, 0.5);
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\draw[color=gray,->] (0.5, 0.0) arc (0:30:0.5);
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\node[right, color=gray] at (0.47, 0.12) {$30^\circ$};
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\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
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\node[above right] at (0.87, 0.5) {$\ket{\psi_0}$};
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\draw[dotted] (0, 0) -- (-0.707, -0.707);
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\draw[color=gray,->] (0.25, 0.0) arc (0:-135:0.25);
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\node[below, color=gray] at (0.2, -0.2) {$135^\circ$};
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\fill[color = ored] (-0.707, -0.707) circle[radius=0.05];
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\node[below left] at (-0.707, -0.707) {$\ket{\psi_1}$};
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\end{tikzpicture}
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\end{center}
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\vfill
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As you may have noticed, we don't need two coordinates to fully define a quibit's state. \par
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We can get by with one coordinate just as well.
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\vspace{2mm}
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Instead of referring to each state using its cartesian coordinates $a$ and $b$, \par
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we can address it using its \textit{polar angle} $\theta$, measured from $\ket{0}$ counterclockwise:
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[dotted] (0, 0) -- (0.707, 0.707);
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\draw[color=gray,->] (0.5, 0.0) arc (0:45:0.5);
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\node[above right, color=gray] at (0.5, 0) {$\theta$};
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\fill[color = ored] (0.707, 0.707) circle[radius=0.05];
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\node[above right] at (0.707, 0.707) {$\ket{\psi}$};
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\end{tikzpicture}
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\end{center}
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\problem{}
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Find $a$ and $b$ in terms of $\theta$ for an arbitrary qubit.
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\vfill
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\pagebreak |